You have found the following ages (in years) of all 5 zebras at your local zoo: $ 13,\enspace 1,\enspace 10,\enspace 37,\enspace 7$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 5 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{13 + 1 + 10 + 37 + 7}{{5}} = {13.6\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $13$ years $-0.6$ years $0.36$ years $^2$ $1$ year $-12.6$ years $158.76$ years $^2$ $10$ years $-3.6$ years $12.96$ years $^2$ $37$ years $23.4$ years $547.56$ years $^2$ $7$ years $-6.6$ years $43.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.36} + {158.76} + {12.96} + {547.56} + {43.56}} {{5}} $ $ {\sigma^2} = \dfrac{{763.2}}{{5}} = {152.64\text{ years}^2} $ The average zebra at the zoo is 13.6 years old. The population variance is 152.64 years $^2$.